JEE Main 2014 — Mathematics Algebra

We know that the ${n}^{th}$nth term and sum of $n$nterms of an $A.P.$A.P. with first term $a$a and common difference $d$d are respectively $a+(n-1)d$a+(n−1)d and $\frac{n}{2}[2a+(n-1)d].$2n​[2a+(n−1)d].

Let, the number of terms in the given $A.P.$A.P. be $2n$2n then there are $\frac{2n}{2}=n$22n​=n even terms and $n$n odd terms.

Then, ${T}_{1}=a$T1​=a and ${T}_{2n}=a+(2n-1)d$T2n​=a+(2n−1)d

Given ${T}_{2n}-{T}_{1}=\frac{21}{2}$T2n​−T1​=221​

$$\Rightarrow (2n-1)d=\frac{21}{2}$$⇒(2n−1)d=221​

$$\Rightarrow 2nd-d=\frac{21}{2}...(i)$$⇒2nd−d=221​...(i)

Also, the sum of odd terms i.e. $a+(a+2d)+(a+4d)+...$a+(a+2d)+(a+4d)+... is

$$\frac{n}{2}[2a+(n-1)2d]=24$$2n​[2a+(n−1)2d]=24

$$\Rightarrow 2a+(n-1)2d=\frac{48}{n}...(ii)$$⇒2a+(n−1)2d=n48​...(ii)

And, the sum of even terms i.e. $(a+d)+(a+3d)+(a+5d)+...$(a+d)+(a+3d)+(a+5d)+... is

$$\frac{n}{2}[2(a+d)+(n-1)2d]=30$$2n​[2(a+d)+(n−1)2d]=30

$$\text{⇒}\frac{n}{2}[2a+(n-1)2d+2d]=30$$⇒2n​[2a+(n−1)2d+2d]=30

Put the value from equation $(ii)$(ii) to get

$$\frac{n}{2}[\frac{48}{n}+2d]=30$$2n​[n48​+2d]=30

$$\Rightarrow 24+dn=30$$⇒24+dn=30

$$\Rightarrow nd=6...(iii)$$⇒nd=6...(iii)

Put this value in the equation $(i),$(i), to get

$$2\times 6-d=\frac{21}{2}$$2×6−d=221​

$$\Rightarrow d=\frac{24-21}{2}=\frac{3}{2}$$⇒d=224−21​=23​

Now, put $d$d in the equation $(iii),$(iii), to get

$$\Rightarrow n\times \frac{3}{2}=6$$⇒n×23​=6

$$\Rightarrow n=4$$⇒n=4

$$\Rightarrow 2n=8$$⇒2n=8

Thus, the number of terms in the given $A.P.$A.P. is $8.$8.